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I have a probably quite simple question RE the HST.

After some work, I obtain as the partition function for the infinite range 1D Ising model $$Z = \int_{-\infty}^\infty \frac{dy}{\sqrt{2\pi / N\beta J}} e^{-N\beta f(y)}$$ where $$f(y)= \frac{J}{2} y^2 - \frac{1}{\beta} \ln[2\cosh(\beta(h + Jy))]$$

Now I want to evaluate the integral in the thermodynamic limit and thus I will be using the method of steepest descent. I am asked to show that $$Z = \sum_i e^{-\beta N f(y_i)}$$ and have to find the equation satisfied by the $y_i$.

So of course my idea was to say that I will get a contribution from each of the local minima of $Z$, which leads to the condition $$\frac{\partial f}{\partial y}|_{y_i} = 0$$ and $$\frac{\partial^2 f}{\partial y^2}_{y_i} > 0$$ However, to obtain the exact form I am asked to show, I'd also need the condition that $$\frac{1}{2} \frac{\partial^2 f}{\partial y^2}|_{y_i} = J$$ because otherwise the prefactors don't come out correct to cancel each other. And now I just don't see how I get this result since the second derivative of $f$ becomes rather messy.

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I must be missing something, but the first derivative gives you the correct equation for the order parameter y, and the second derivative is a constant, regardless of its value (which becomes a multiplicative constant to the partition function, which is nothing more than a normalization -- it can't affect any thermodynamics) – wsc Oct 13 '11 at 5:00
But the factor contains some $\beta$ and thus $T$-dependence? – Lagerbaer Oct 13 '11 at 5:26
I haven't done the complete calculation (no CAS available...), just an idea. The condition $f'(y_i) = 0$ leads to a transcendental equation $ y_i = \tanh(\beta(h + Jy_i))$. For the second derivative , you use $[\tanh]' = 1/\mathrm{sech}^2 = 1 - [\tanh]^2$, which you can relate to $y_i$ by the above relation. Have you done this? – Gerben Oct 13 '11 at 16:36
Well okay, this gives $f''(y)|_{y_i} = J - J^2 \beta (1 - y_i^2)$. Not sure how to go from there. – Lagerbaer Oct 13 '11 at 18:16
More on Hubbard-Stratonovich-transformation, thermodynamic limit and steepest descent: now – Qmechanic Feb 23 '12 at 13:33

"But you do not need such sophistication to solve this homework problem". Pfft.

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The link is broken... – kηives Feb 17 '13 at 17:32

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